- This topic has 5 replies, 3 voices, and was last updated 4 years ago by
.
- Posts
Question 1
A company needs 400kg of material Z to fulfil a customer order in one month’s time
It currently has no material Z in inventory. The current purchase price of material Z is 20$ per kg and this is expected to rise to 24$ per kg in one month’s time. Material Z is perishable and normally 20% of stored material is lost per month
The company expects to have 200kg of material Y in inventory in one month’s time with no alternative use other than to sell it for scrap fot 18$ per kg. The 200kg of material Y could be converted into 200kg of material Z in one month’s time at a cost of 4$ per kg.Material Y is not perishable.
What is the total relevant cost of Material Z to fulfill the customer order?
Material Z is is regular use Probablly
Here is Material Y probably was bought in the past and no can be used
Material Y has not alternative use but can be converted to Material Z1) Conversion of Material Y to Material Z (18$ scrap value+4$ modification cost)=22
2)Buy Material Z in one month’s time-24$ per kg
3)Buy Material Z and store it for one month time–20$/0.8=25Question 2
UU company has been asked to quote fot a special contract.The following information about the material needed has been given
Material X
Book valuie————–Scrap value———————-Replacement cost
5$ per kg——————–0.50$ per kg—————————5.50$ per kgThe contract requires 10Kgs of Material X. There are 250kgs of this material in inventory which was purchased in error over two years ago. If Material X is modified ,at a cost of 2$ per kg, it could then be used as a substitute for Material Y which is in regular use and currently costs 6$ per Kg
What is the relevant cost of the materials for the special contracts?
Material Y is is regular use
Here is Material X probably was bought in the past and no can be used
Material X has not alternative use but can be converted to Material Y1)conversion of Material X to Material Y (scrap value 0.50+2modification cost)=2.50
2)Buy material Y-6$
3)6-2=4-let us compare the two example
why i deduct 2$ of modification cost of X from material y’s current price of 6
why not in the follwing
2)Buy Material Z in one month’s time-24$ per kg then deduct $4 of modification cost of \\y from material Z such as 24-4=20
3)Buy Material Z and store it for one month time–20$/0.8=25 then deduct 4$=21$ as above
If you did as you have written at the end in options 2 and 3 (and subtracted the $4), then when comparing the costs with option 1 you would not in the comparison have added the $4 to the scrap value (because that would be accounting for it twice). So you would still choose option 1 as the cheapest. This would just be for deciding which was the best option.
Having chose option 1, then the actual cost involved would involve paying the $4 and the relevant cost would be $22.
1)conversion of Material X to Material Y (scrap value 0.50+2modification cost)=2.50
2)Buy material Y-6$
3)6-2=4Buying material Y at 6$ is not relevant for the company it costs too much so we skip this option
Conversion 2+0.50=2.50
NRV=6-2=4
Minimum 0.50in the answer the comparison is made between 0.50 and 4 why not 2 can not be added over 0.50 as in the second example?I know that in opportunution cost if there is alternative use then opportunity cost is chosen accoording in higher value
in the second question
1) Conversion of Material Y to Material Z (18$ scrap value+4$ modification cost)=22-here is did not understand as then if it is the case why 0.50+2=2.50 in the first question can not be ?
here is I do not understand because why in the first question 2 is deducted from 6 which is similar situation below
2)Buy Material Z in one month’s time-24$ per kg-4=20
3)Buy Material Z and store it for one month time–20$/0.8=25-4=21Dear tutor, you probably understood why i stuck in the second question
I usually solve this kind of question by deducting any modification cost of material from substitute product’s selling price to get NRV and compare it with scrap value and seleect the higher between them like in the first exampleI deduct 6-2=4Nrv and scrap value and select in the higher one but in the second question first time i saw i add modification cost over scrap value if this is the case why in in the first example?
In tha case of question 1, they will convert the 200 kg of Y into Z and so the cost per kg is 18 + 2 = $20. The other 200 kg needed will have to be purchases and will cost $24 per kg
Question 2 is different because it is material X that is required (and not material Y).
If they did not need X for the contract they could either sell it and get $0.50 or could convert it to Z at a cost of $2 and save $6 because the would not need to buy some Z. They would prefer to convert is and save a net $4.
If they use it in the contract then they will not be able to save $4 and so the relevant cost is $4.You cannot compare the two questions because in Q1 it is material Z that is needed for the contract. In Q2 it is material X that is needed for the contract.
In question 1;
1) if we purchase now we will spend $20*400=8000. lost in a month time(perishable) is 20% of 400=80kgs. which we will have to bought at the mkt value of the time. 80*24=1920. total is therefore 8000+1920=9920.
2) if we decide to buy after a month(when the item is needed) then we will spend,24*400=9600.
3) if we convert Y to Z it will cost us $4 per kg. that will save us $20 per kg of getting material Z at the mkt value($24). 200*20=4000. the remaining 200kgs will have to be purchased therefore;200*24=4800 total is 8800.
the company consider the most economical and therefore relevant cost is 8800.This is how i understood this question. is this the right way?
thanks.
- The topic ‘relevant cost’ is closed to new replies.